A spring (k=75N/m) has an equilibrium length of 1.00m. The spring is compressed

Question

A spring (k=75N/m) has an equilibrium length of 1.00m. The spring is compressed to a length of 0.50m and a mass of 2.0kg is placed at its free end on a frictionless slope which makes an angle of 41 degrees with respect to the horizontal. The spring is then released. (a) If the mass is not attached to the spring, how far up the slope will the mass move before coming to rest? (b) if the mass is attached to the spring, how far up the slope will the mass move before coming to rest? (c) Now the incline has a coefficient of kinetic friction k. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction k?

Explanation / Answer

the energy of a compressed spring = kx^2/2....x= compression. as no friction acts ,we can conserve energy,let h be the ht to which the block rises,and d be the dis i covers on slpoe. so d=h cosec(41). kx^2/2=mgh=mgdsin(41) so , calculate d=k (0.5)^2/2mgsin(41) =1.458m force = kx and gravity force=mgsin(41)...vertical component of g. so,again conserving energy, (kx + mgsin(41))d` =mgh`....work = energy = force x displacement. both spring and gravity oppose the motion of block. now put the values and calculate urself...again the trigo relation remains same. now when there is friction it`s value is umgcos(41)....as the normal force ids mgcos(41) ...componennt of g along contact force. so , as the block stops when spring is at it`s equilibrium length, equqting the work done by friction to that energy lost by spring . umgcos(41) x 0.5 = k(0.5)^2/2 you can calculate the value.

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